# implicit differentiation easy question that i can't do...

• Jan 3rd 2010, 06:08 PM
differentiate
implicit differentiation easy question that i can't do...
Find the points on the graph $x^3 + y^3 = xy$
that are parallel to the x axis (gradient = 0)

so I have to differentiate it
$3x^2 + 3y^2 (dy/dx) = dy/dx$

then what?

Thank you
• Jan 3rd 2010, 06:20 PM
skeeter
Quote:

Originally Posted by differentiate
Find the points on the graph $x^3 + y^3 = xy$
that are parallel to the x axis (gradient = 0)

so I have to differentiate it
$3x^2 + 3y^2 (dy/dx) = dy/dx$

correction, product rule on the right hand side ...

$\textcolor{red}{3x^2 + 3y^2 \frac{dy}{dx} = x \frac{dy}{dx} + y}$

then what? solve for $\textcolor{red}{\frac{dy}{dx}}$ and set the derivative equal to 0. use that equation to find a relationship between y and x to find the points on the curve where the gradient is 0.

...
• Jan 3rd 2010, 07:42 PM
differentiate
thank you
so then,

$3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y$
$3y^2(dy/dx) - x(dy/dx) = y - 3x^2$
$(dy/dx)(3y^2 - x) = y - 3x^2$
$dy/dx = (y-3x^2)/(3y^2 - x)$
dy/dx = 0
$y - 3x^2 = 0$
$y = 3x^2$

do I put that into the original equation?

$x^3 + (3x^2)^3 = x(3x^2)$
$x^3 + 27x^6 = 3x^3$
$27x^6 - 2x^3 = 0$
$x^3(27x^3 - 2) = 0$

I'm stuck
and I don't have answers to this question
• Jan 3rd 2010, 07:56 PM
dedust
Quote:

Originally Posted by differentiate
thank you
so then,

$3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y$
$3y^2(dy/dx) - x(dy/dx) = y - 3x^2$
$(dy/dx)(3y^2 - x) = y - 3x^2$
$dy/dx = (y-3x^2)/(3y^2 - x)$
dy/dx = 0
$y - 3x^2 = 0$
$y = 3x^2$

do I put that into the original equation?

$x^3 + (3x^2)^3 = x(3x^2)$
$x^3 + 27x^6 = 3x^3$
$27x^6 - 2x^3 = 0$
$x^3(27x^3 - 2) = 0$

I'm stuck
and I don't have answers to this question

then the points are
$x = 0$ $\rightarrow y = 0$
$x = \frac{\sqrt[3]2}{3}$ $\rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$
• Jan 3rd 2010, 08:01 PM
pickslides
Quote:

Originally Posted by dedust
then the points are
$x = 0$ $\rightarrow y = 0$
$x = \frac{\sqrt[3]2}{3}$ $\rightarrow y = \left(\frac{\sqrt[3]2}{3}\right)^2$

You mean $x = \frac{\sqrt[3]2}{3}$ $\rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$
• Jan 3rd 2010, 08:15 PM
dedust
Quote:

Originally Posted by pickslides
You mean $x = \frac{\sqrt[3]2}{3}$ $\rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2$

oops,..yeap
(Rofl)